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Homework Help: Newton's Law of Temperature Change Differential

  1. Oct 11, 2012 #1
    I'm a bit confused as to what I do next with this problem.

    Consider the initial value problem [itex]\frac{dy}{dx}[/itex] = a(y-b) where y(0) = y0

    With a > 0, b = 0, it represents exponential growth, while a < 0, b = 0 gives exponential decay. With y = T, a = -k, k = TA and y0=T0 it gives Newton's Law of Temperature Change,

    [itex]\frac{dy}{dx}[/itex] = -k(T - TA)

    Solve the given model and give a qualitative sketch for each of the cases 1) a > 0 and 2) a < 0. You may assume b ≥ 0. Include y0= b as one typical solution for each.

    So what I did was I applied the integral to both sides of [itex]\frac{dy}{dx}[/itex] = -k(T - TA) yielding,

    1 = -k ∫ (T - TA)

    But I don't know how to integrate TA
    Last edited: Oct 11, 2012
  2. jcsd
  3. Oct 11, 2012 #2


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    You can't integrate both sides because you have no differential on the right. Part of your problem is that you say y= T but then leave y in the equation.
    Write the equation as [itex]\frac{dT}{dx}= -k(T- T_A)[/itex], then as
    [tex]\frac{dT}{T- T_A}= -kdx[/tex]
    and integrate. TA is a constant. To integrate on the left, let u= T- TA so that du= dT.
  4. Oct 11, 2012 #3

    Ray Vickson

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    I think you are mis-reading the problem. It says to solve the given model, and to look at the cases a > 0 and a < 0, etc. To me, that says solve dy/dx = a*(y-b) [= the given model], so there is no "T" involved.

  5. Oct 11, 2012 #4
    Oh, so we would integrate the constant b as a normal constant correct? I.e. b*x?
  6. Oct 11, 2012 #5


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    Not when it is in the denominator! Use the substitution u= T- TA as I suggested.
  7. Oct 11, 2012 #6
    Oh I understand, thank you!
  8. Oct 11, 2012 #7
    Wait, it's dT/dt not dT/dx?
  9. Oct 11, 2012 #8
    The only thing I'm not sure of is if it the question is asking for solutions. If it is can I put the dy/dx equation in the standard differential form,

    dy/dx + P(x) = Q(x)

    Then, I can determine the integrating factor and solve that way. I was initially thinking of doing that, would it work?
    Last edited: Oct 11, 2012
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